Math destroyer test 1 question 6

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
it says the total time for the trip is thus 21/(x+2)+21/(x+2) = 10
idk what he mean by this im very new to destroyer.

thanks

Don't have destroyer...

But distance = speed/time.
So time = speed/distance

total time must have been 10, while the speed of both objects must be 21, while the distance was x+2.

time(1) + time(2) = time(total)
speed(1)/distance(1) + speed(2)/distance(2) = time(total)

hope that helps!
 
Don't have destroyer...

But distance = speed/time.
So time = speed/distance

total time must have been 10, while the speed of both objects must be 21, while the distance was x+2.

time(1) + time(2) = time(total)
speed(1)/distance(1) + speed(2)/distance(2) = time(total)

hope that helps!

distance should be speed x time
time should be distance/speed
 
Does anyone know what to do after getting :

21/(x+2) + 21/(x-2) = 10

Take the inverse of both sides and you get a common denominator (21).

[(x+2)+(x-2)]/21 = 1/10

Always focus on getting all variables in the numerator. it makes life easier.

The other way to think about it is to multiply the left side by (x+2)/(x+2) and (x-2)/(x-2) - since you're multiplying by 1 you don't have to do both sides of the equation. You're just rearranging it so all variable terms are on the top.
 
Last edited:
Take the inverse of both sides and you get a common denominator (21).

[(x+2)+(x-2)]/21 = 1/10


Always focus on getting all variables in the numerator. it makes life easier.

The other way to think about it is to multiply the left side by (x+2)/(x+2) and (x-2)/(x-2) - since you're multiplying by 1 you don't have to do both sides of the equation. You're just rearranging it so all variable terms are on the top.

That's straight-up voo-doo algebra bud...you CANNOT do that.

Here, just do it the long way...common denominator and then rearrange:

21/(x+2) * (x-2)/(x-2) + 21/(x-2) * (x+2)/(x+2) = 10
21(x-2) + 21(x+2) = 10(x+2)(x-2)
21x-42 + 21x+42 = 10(x^2-4)
0 = 10x^2 - 42x - 40
0 = 5x^2 - 21x - 20
0 = (5x + 4)(x - 5)

only positive solution is x = 5
 
So I've tried doing this problem again for the second time going through destroyer. And while I understand the solution to the back of the book (at least I think) and how they got an answer of 3, why can't you do this:

d=rt
total distance--> 42=[(x+2)+(x-2)]10
and once you solve for x, plug into 21=(x+2) t and solve for t?
I did it that way and got an answer of 5 hrs. I can't seem to figure out why this method doesn't give the answer of 3 hrs. Any help?
 
So I've tried doing this problem again for the second time going through destroyer. And while I understand the solution to the back of the book (at least I think) and how they got an answer of 3, why can't you do this:

d=rt
total distance--> 42=[(x+2)+(x-2)]10
and once you solve for x, plug into 21=(x+2) t and solve for t?
I did it that way and got an answer of 5 hrs. I can't seem to figure out why this method doesn't give the answer of 3 hrs. Any help?
The x that you solved for represents the speed of the boat not time.
To find time you do 21/(5+2) = 3 hrs
The way i did this problem I didn't bother solving for x at all
Going down stream : 21= t1(x+2)
going up stream : 21= t2(x-2)
which means that 21 is a multiple of both t1 and t2 the only valid answers are 3 and 7 and since we want the time it takes to go downstream it's 3

Hope this helps
 
The x that you solved for represents the speed of the boat not time.
To find time you do 21/(5+2) = 3 hrs
The way i did this problem I didn't bother solving for x at all
Going down stream : 21= t1(x+2)
going up stream : 21= t2(x-2)
which means that 21 is a multiple of both t1 and t2 the only valid answers are 3 and 7 and since we want the time it takes to go downstream it's 3

Hope this helps
Thank you! That shortcut is nice. I hope I remember it can be done that way if i encounter a similar problem. 👍
 
Top